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ira [324]
3 years ago
15

A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The

concentration of CaCl2 in this solution is ________ molal. A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is ________ molal. 5.70 0.214 0.569 63.2 1.76
Chemistry
1 answer:
polet [3.4K]3 years ago
3 0

Answer:

The concentration of CaCl2 in this solution is 0.563 M

Explanation:

<u>Step 1:</u> Data given

mass of CaCl2 = 23.7 grams

Molar mass of CaCl2 = 110.98 g/mol

mass of water = 375 grams

Molar mass of water = 18.02 grams

Volume of the solution = V

<u>Step 2:</u> Calculate the mass of solution

Mass of the solution = Mass of solute + Mass of solvent

Mass of the solution = 23.7 grams + 375 grams

Mass of the solution = 398.7 grams

<u>Step 3:</u> Calculate the volume

Volume = mass / density

Volume = 398.7 grams / 1.05 mg/mL

Volume = 379.71 mL = 0.3791 L

<u>Step 4:</u> Calculate moles of CaCl2

Number of moles CaCl2 = mass CaCl2 / Molar mass CaCl2

Number of moles CaCl2 = 23.7 grams / 110.98 g/mol

Number of moles CaCl2 = 0.21355 moles

<u>Step 5:</u> Calculate the concentration of CaCl2

Concentration = Number of moles / Volume

Concentration = 0.21355 moles / 0.3791 L

Concentration = 0.563 M

The concentration of CaCl2 in this solution is 0.563 M

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crimeas [40]

Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

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                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

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b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

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The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

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